Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct abc {
  int rc, x;
} st[100000];
int k = 0, ans = 0;
int main() {
  ios_base::sync_with_stdio(false);
  int n, m, C = 0;
  cin >> n >> m;
  int a[n][m];
  if (n < m) {
    int i, j, A, B;
    for (i = 0; i < n; i++) {
      A = INT_MAX;
      for (j = 0; j < m; j++) {
        cin >> a[i][j];
        if (a[i][j]) {
          C = 1;
        }
        if (a[i][j] < A) {
          A = a[i][j];
        }
        if (i > 0) {
          a[i - 1][j] -= B;
        }
      }
      ans += A;
      int kk = k + A;
      for (; k < kk; k++) {
        st[k].rc = 1;
        st[k].x = i + 1;
      }
      B = A;
    }
    for (j = 0; j < m; j++) {
      a[n - 1][j] -= B;
    }
    int ca, cb;
    for (int c = 0; c < m; c++) {
      ca = INT_MAX;
      for (int r = 0; r < n; r++) {
        if (a[r][c] < ca) {
          ca = a[r][c];
        }
        if (c > 0) {
          a[r][c - 1] -= cb;
        }
      }
      ans += ca;
      int kk = k + ca;
      for (; k < kk; k++) {
        st[k].rc = 2;
        st[k].x = c + 1;
      }
      cb = ca;
    }
    for (int r = 0; r < n; r++) {
      a[r][m - 1] -= cb;
    }
  } else {
    int i, j;
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        cin >> a[i][j];
        if (a[i][j]) C = 1;
      }
    }
    int ca, cb;
    for (int c = 0; c < m; c++) {
      ca = INT_MAX;
      for (int r = 0; r < n; r++) {
        if (a[r][c] < ca) {
          ca = a[r][c];
        }
        if (c > 0) {
          a[r][c - 1] -= cb;
        }
      }
      ans += ca;
      int kk = k + ca;
      for (; k < kk; k++) {
        st[k].rc = 2;
        st[k].x = c + 1;
      }
      cb = ca;
    }
    for (int r = 0; r < n; r++) {
      a[r][m - 1] -= cb;
    }
    int ra, rb;
    for (int r = 0; r < n; r++) {
      ra = INT_MAX;
      for (int c = 0; c < m; c++) {
        if (a[r][c] < ra) {
          ra = a[r][c];
        }
        if (r > 0) {
          a[r - 1][c] -= rb;
        }
      }
      ans += ra;
      int kk = k + ra;
      for (; k < kk; k++) {
        st[k].rc = 1;
        st[k].x = r + 1;
      }
      rb = ra;
    }
    for (int c = 0; c < m; c++) {
      a[n - 1][c] -= rb;
    }
  }
  int flag = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (a[i][j] != 0) {
        flag = 1;
        break;
      }
    }
    if (flag) break;
  }
  if (C) {
    if (flag == 1) {
      cout << -1 << endl;
    } else {
      cout << ans << endl;
      for (int i = 0; i < k; i++) {
        if (st[i].rc == 1)
          cout << "row ";
        else
          cout << "col ";
        cout << st[i].x << endl;
      }
    }
  } else {
    cout << 0 << endl;
  }
  return 0;
}